First slide

Vector triple product

Question

If $\vec{a}$ is parallel to $\vec{b} \times \vec{c}, then (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c})$ is equal to

Moderate

A

$| \vec{a} |^{2} (\vec{b} \cdot \vec{c})$
B

$| \vec{b} |^{2} (\vec{a} \cdot \vec{c})$
C

$| \vec{c} |^{2} (\vec{a} \cdot \vec{b})$
D

none of these

Solution

$\begin{array}{l} (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{u}, where \vec{u} = \vec{a} \times \vec{c} \\ \Rightarrow \vec{a} \cdot (\vec{b} \times \vec{u}) = \vec{a} \cdot [\vec{b} \times (\vec{a} \times \vec{c})] \\ = \vec{a} \cdot [(\vec{b} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{b}) \vec{c}] \\ = \vec{a} \cdot (\vec{b} \cdot \vec{c}) \vec{a} (∵ \vec{a} \cdot \vec{b} = 0) \\ = | \vec{a} |^{2} (\vec{b} \cdot \vec{c}) \end{array}$

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